Question

1. Construct an active high pass RL filter circuit with the following requirements: • L = 100mH • Cutoff frequency = 5500 Hz Gain = 5V/V • Conduct an AC Sweep & indicate the cutoff frequency on the graph using curs ursors

Answer 1

Solution: Using the transfer function model of voltage gain of active high pass filter and using the expressions of

passband gain and cutoff frequency relations the remaining components values are easily obtained as derived below -

Solution : To Here in design * от Арман R-1 вие Design specifications w R o vout 2=10omh fe Gain Ava suv cue.off frequency: 5500hz Vir Transforming the circuit into s-domain ainder across 'Si' Using vostage vouto) branch SSL si VA = Vinco x 34 (R734) SL -X Vints) (R4L) Due to virtualground of op. Amp - v= pt inverning input terminal Apply Kel at "[***] - Umut = 0 - Put the value of v in equ♡ SL Vinis) Voutis) (***) (Q+54) RE Voutis) Rf y SL 1+1 Vinis) (atse) R RA

-> M(S) Voutis) Vents) SL (1) (R +SL). Here Ap= It If is passband gain of the filter As sejo x Af Hejw) = Voutliw) jul kinijos) (Rijal) Itejol=0 4 W +0 AS W100 Hijooll =lim WL GOL x Af x AF with TR + (we)? W R? +L 2 L X AF { x 44 = AR (+2) = 0 and the jar / 70 This shows the giren circuit acts as high pass filter News, to attain out of frequenys tres equated of equating the At, lejo) = magnitude of trijos) = 1 / Hemax at wale maline? where 'Hmax = Af' is pass Band goins of frier 'as is cut-off frequency in oud/sove Лую, | Htv9) w.xAf JR 40. At w = Weisec A = AX COL 52 JR? + (602) ? | w=weatsee Squasing both sides

2 Wc L Wel 2+144.2) => 220L%* = RPH (291) CL) ² = R2 =u? Э 4 & radisec cut-off frequency in Hesta fc = We Now, for achieving te= 5soothe Gain = sulu svv L=100mm value of 'R' required & Ro QX Life or R = 2R x 100x70%x5500 [R = 3.4559 kr tha Now, Gain = 5v/v 7 Af = 5=14 be RF = 54 4K2 IK = 4 R, Rf = 4R Set | Ri=1k2 Rf = 46.2 therefore, final circuit required - Using e verify calculations, Vo 3.4557K-2 w the MULTISIM we can 100ml Answer.

R2 4k R1 vout 1kQ U1 R3 vin 3.4557k0 OPAMP_3T_VIRTUAL V1 1Vpk 1kHz 0° L1 100mH H Circuit simulation using Multisim

х Analyses and Simulation Active Analysis: Interactive Simulation AC Sweep DC Operating Point Frequency parameters Output Analysis options Summary AC Sweep 1 Transient Hz Reset to default 1. 100 kHz DC Sweep Start frequency (FSTART): Stop frequency (FSTOP): Sweep type: Number of points: Vertical scale: Linear Analyses and Simulation х Single Frequency AC 1000 Active Analysis: Parameter Sweep Linear 3. Noise AC Sweep Interactive Simulation Monte Carlo 2. DC Operating Point Fourier AC Sweep Selected variables for analysis: All variables Vivout) (vin) Temperature Sweep Transient Distortion DC Sweep Single Frequency AC Sensitivity 5. Frequency parameters Output Analysis options Summary Variables in circuit: All variables I(L1) I(R1) I(R2) I(R3) I(V1) P(L1) P( R1) Add P(R2) P(R3) P(V1) Remove V (1) V(3) Edit expression... V(vin) Vivout) 4. Add expression. Worst Case Parameter Sweep Noise Noise Figure Add this expression to plot the Frequency response curve. Pole Zero Monte Carlo Transfer Function Fourier Trace Width Temperature Sweep Batched Distortion User-Defined Sensitivity 6. Run Save Cancel Help A.C Simulation settings

Design 1 AC Sweep 5- 11 Cursor V (vout)/V (vin) Magnitude (5.5000k, 3.5354) Cut-off frequency (in kHz) f x 1 yi x2 y2 44 dx dy dy/dx 1/dx 5.5000k 3.5342 100.0000k 4.9992 94.5038 1.4581 15.4286 10.5816 Pass Band Gain (at high frequency)= 4.9992 - 5VN Magnitude becomes 0.707 times of passband gain at fc =5.5kHz Ok 20k 40k 60k 80k 100K Frequency (Hz) Vivout)/V(vin) Magnitude Plot obtained using MULTISIM